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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

You'd be an asset on the debating team, Bobby!

Surely the algebraic method is better for non-integer solutions, particularly those with multiple decimal places?

For me, the grass is greener over your way...

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

Hi phrontister;

Uh, oh! I feel a rant coming on. Yep, here it comes.

Rant:

I don't agree that the algebraic way is necessarily better. The numerical methods work on all types of equations and problems. They are very robust. Algebraic methods, look good but work less often. Generally, they only work on book type problems. They are usually hundreds of years old and invented by guys who were forced to do math with quill pens and candles.

The numerical methods typify what the Borweins call experimental math. They are reviving how Gauss, Newton and Euler preferred to work, experimentally. The emphasis is on numerics, computation, computers and less on proof. More discrete, less continuous math is their motto.

So if you had a bunch of methods that worked on 11 problems and made the guys at the AOPS forum tremendously happy as opposed to methods that did em all which would you prefer?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi Bobby,

I'm way out of my depth here and can't comment much...except to say that each method probably has strengths and weaknesses that the other doesn't have, and one might suit a particular application better than the other. In what ratio that is I wouldn't have a clue.

And they can work together, as in a 'crossed ladders' problem I came across many years ago (see diagram). I came up with an equation in terms of x but couldn't solve it, and then used Newton's method of successive approximations to find the value of x.

I went to the library to try to find out how to solve nasty equations and read about Newton's method, but then had to learn up on it and try to understand what 'functions' and 'derivatives' were because I didn't get to that level at school.

Eventually I worked out how to program that into a little hand-held computer I've got (a Sharp PC-1500A), which automated the solving process.

That was all very interesting at the time...but now I've forgotten it all.

I've since found lots of discussion about this problem, and similar, on the internet. My equation doesn't seem to be on the sites I looked at...but maybe it's there, in a disguise I missed.

*Last edited by phrontister (2010-02-02 16:04:05)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

Hi;

I had a bunch of Sharp programmable calculators. Last one had 5k of memory and a larger screen. Seems like a hundred years ago.

Take for instance a poster who was asking how to solve a cubic. I was going to show him great ideas in iteration that would have knocked the socks off that cubic. Instead, because his school training was so poor I ended up directing him to that horrible algebraic idea that was invented in the Renaissance by Tartaglia and stolen by Cardano. Don't get me wrong, both of those men were incredible geniuses but the method is ancient and awful.

My point, we have computers, we have the internet let's use them.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi Bobby,

My point, we have computers, we have the internet let's use them.

I do. I like using BASIC, Excel and other solvers where I can, but I do spark up if I happen to see a nifty algebraic solution.

I had a bunch of Sharp programmable calculators. Last one had 5k of memory and a larger screen.

My Sharp PC-1500A has an 8kB memory, which I tripled to a whopping-great 24kB by adding a 16kB module to it years ago. I still enjoy using it.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

24kb of memory!!! I had an old TRS-80 mark 3 computer that had a whopping 16k of memory, a Z-80 chip and it came with it's own BW screen and TRS-80 Micro$oft Basic built in ROM. Didn't even have a floppy disk, you used a cassette player at an incredible 1500 baud.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

It's amazing how many programs could be squeezed into those old machines. Helped me to develop bad program-writing habits, though, by jamming in the max amount of code per line just to save enough kBs here and there to make room for another program.

I used to record my PC-1500A's programs on cassette via a radio-cassette player. Had to twiddle the knobs and tweak the settings to get them just right for a successful upload...and download!

Btw, my equation for the crossed ladder problem was 1/√(16-x²) + 1/√(36-x²) - 1 = 0...which my Sharp 48SX solves easily with its Solver function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

Helped me to develop bad program-writing habits, though

No it doesn't. Teaches you how to write tight code. Not like the bloated gaaaarbaaagge I see today.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

This one is so easy, I expect 414 answers.

Problem #21:

A basketball player has gone 21 for 47 from the foul line for a poor percentage of .4468. In order not to be cut from the team he must get his foul shooting percentage to 70 % in his next 200 free throw attempts. What is the minimum number of free throws he must he make out of his next 200 attempts to get his average to 70%. Round your answer to the nearest free throw and please hide it.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi Bobby,

I reckon the basketballer has to make at least

many free throws not to be cut from the team.I expect 414 answers

Why 414?

*Last edited by phrontister (2010-02-02 22:56:25)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

Why 414?

Because it is so easy.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

But MIF has *many *more members than 414...not to mention guests.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

True. But that doesn't change the fact that 414 is the number of responses I am expecting.

Nice answer. You solved it simply, I solved it in a hideous way on some other forum and they are going to shred me for my ugly solution.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

These type are always a pain.

Problem #22:

A*(ELVIS) + PRESLEY + 1935 + 1977 + P = PPPPPPP

Each letter stands for a different number from 0 to 9. You cannot have a leading digit of 0. How many solutions are there? Can you find any?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I agree that numerical methods are usually quicker and easier to use, but I still think that it's best to be algebraic whenever possible.

It's just so much nicer to be *absolutely correct* than just very accurate. Numerical methods are fine for any and all practical purposes, but to get a closed-form expression is much more pleasing.

Why did the vector cross the road?

It wanted to be normal.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi Bobby,

I had a go at the King's puzzle last night, but got stuck.

is where I'm up to so far. I've probably done something wrong, though, because I should be able to finish it from there, but can't...unless I've overlooked something. I'll try again tonight.*EDIT:* Just saw that I made a silly, flyspot-sized error that I'm sure will make a huge difference. Start again.

*Last edited by phrontister (2010-02-03 11:58:35)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

Hi phrontister;

How many total integer solutions are there to the equations?

a + b + c + d + e + f = r

with r = 0,1,2,3,4,...60

f >= e >= d >= c >= b >= a >= 0

a,b,c,d,e,f < 11

phrontister wrote:

I wondered also if it could somehow be done with permutations, but I know nothing about that subject.

There are 3 main ways to solve a combinatorics problem (typified by the phrase how many ways).

1) Split it into smaller sub problems that are easier to solve. In programming this is called top down design.

2) Look at a couple of examples of smaller but similar problems and spot the pattern. Then prove the pattern applies for the larger problem.

3) Bijection - turn the problem into a totally different one that has the same number of ways as the original one and is solvable.

The third method is used here: The above problem is translated from the number of integer solutions of a diophantine equation to a block walking experiment. See the drawing. The question now is how many paths are there from the bottom left corner to the top right corner. If you can only go right or up. I have shown one path.

These are easy to solve both combinatorically. You are trying to go from (0,0) to (6,10) which has

total ways. This result is from the BMO.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi Bobby,

is my answer to the King's puzzle. I enjoyed that...hope the answer is correct.

I'll have a closer look at your permutations post when I'm more awake.

*Last edited by phrontister (2010-02-04 04:36:05)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

Hi phrontister;

Correct, Very good!

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi Bobby,

I solved the King's puzzle in BASIC (LibertyBASIC).

my code.

*Last edited by phrontister (2010-02-09 04:00:13)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

Hi;

Problem #23:

Moderately tough.

Prove that 5x + 7y + 11z = r has no solutions for r = 13.

Extra credit: Prove that r = 13 is the highest number without solutions.

Hi phrontister;

Thanks for providing your code for the Elvis problem.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

Easy one.

Problem #24:

No leading zeros, and each letter stands for a different digit, Your solution must fit for both equations.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi Bobby,

Thanks for providing your code for the Elvis problem.

I made some changes to it that reduced the run time down to 15 seconds (previously 1 minute).

There was some duplication and other stuff, such as a lot of unnecessary calculating that I eliminated by repositioning some if/then lines.

I've altered the code in my post #120.

*Last edited by phrontister (2010-02-09 04:19:20)*

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**d1whohatesmath****Member**- Registered: 2010-02-08
- Posts: 26

bobbym wrote:

Hi;

This one is moderately hard.

How many permutations of 3 letters can you make out of the word:

clandestine?

can,tie,and,lad(old fashioned word),eat,set,sit,sat,ace,aid,act,ads,

ail,ain(hebrew),ais(3toed sloth),ait,ale,als,alt,ane,ant,ani,ars,art,

arc,ate,CAD,CAT,CEL,DAN,DAL,DEL,DEN,DIE,DIN,DIs,DIT,eds,ELD,

ELS,END,ENS,ETA.

there are many more,but my fingers are getting cramps.heh.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,710

Hi Omaima;

How many permutations of 3 letters can you make out of the word:

That's true for actual words but their are many more answers. The answer does not have to be an actual word. For instance cld and sec and sce are also answers. And many more.

A permutation is a rearrangement of a set of objects or values.

For instance their are six permutations of a,b,c.

a b c

a c b

b a c

b c a

c a b

c b a

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**